To solve the problem of finding the smallest number that, when divided by 20, 25, 35, and 40, leaves specific remainders (14, 19, 29, and 34 respectively), follow these steps:
Identifying the Problem
The number we are looking for can be expressed in terms of the divisors and their respective remainders:
- Let the number be \( N \).
- \( N \equiv 14 \, (\text{mod} \, 20) \)
- \( N \equiv 19 \, (\text{mod} \, 25) \)
- \( N \equiv 29 \, (\text{mod} \, 35) \)
- \( N \equiv 34 \, (\text{mod} \, 40) \)
Transforming the Conditions
To simplify, we can convert these conditions into a consistent format by subtracting the remainders from each divisor:
- \( N + 6 \equiv 0 \, (\text{mod} \, 20) \)
- \( N + 6 \equiv 0 \, (\text{mod} \, 25) \)
- \( N + 6 \equiv 0 \, (\text{mod} \, 35) \)
- \( N + 6 \equiv 0 \, (\text{mod} \, 40) \)
This means we need to find a common multiple of these moduli.
Finding the Least Common Multiple (LCM)
Now, we will calculate the LCM of 20, 25, 35, and 40:
- The prime factorization is:
- \( 20 = 2^2 \times 5 \)
- \( 25 = 5^2 \)
- \( 35 = 5 \times 7 \)
- \( 40 = 2^3 \times 5 \)
- The LCM is calculated as:
- \( LCM = 2^3 \times 5^2 \times 7 = 1400 \)
Calculating the Smallest Number
Now, since \( N + 6 \equiv 0 \), we have:
- \( N + 6 = 1400k \) for some integer \( k \)
To find the smallest positive \( N \):
- Let \( k = 1 \):
- \( N + 6 = 1400 \)
- \( N = 1400 - 6 = 1394 \)
Thus, the smallest number is:
Final Answer
- The answer is \( \textbf{a) 1394} \).